If (2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + ldot | vedclass

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(D) Given the expansion $(2-5x)^{-1/5} = 2^{-1/5} (1 - \frac{5}{2}x)^{-1/5}$.Using the binomial expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \

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$\frac{2}{3}$

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(D) Given the expansion $(2-5x)^{-1/5} = 2^{-1/5} (1 - \frac{5}{2}x)^{-1/5}$.Using the binomial expansion $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \ldots$ where $y = -\frac{5}{2}x$ and $n = -\frac{1}{5}$:$(1 - \frac{5}{2}x)^{-1/5} = 1 + (-\frac{1}{5})(-\frac{5}{2}x) + \frac{(-\frac{1}{5})(-\frac{1}{5}-1)}{2!}(-\frac{5}{2}x)^2 + \ldots$$= 1 + \frac{1}{2}x + \frac{(-\frac{1}{5})(-\frac{6}{5})}{2} (\frac{25}{4}x^2) + \ldots$$= 1 + \frac{1}{2}x + \frac{3}{25} \cdot \frac{25}{8}x^2 + \ldots = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \ldots$Thus,$(2-5x)^{-1/5} = 2^{-1/5} (1 + \frac{1}{2}x + \frac{3}{8}x^2 + \ldots) = 2^{-1/5} + \frac{1}{2} \cdot 2^{-1/5}x + \frac{3}{8} \cdot 2^{-1/5}x^2 + \ldots$Comparing coefficients,$a_1 = \frac{1}{2} \cdot 2^{-1/5}$ and $a_2 = \frac{3}{8} \cdot 2^{-1/5}$.Therefore,$\frac{a_1}{a_2} = \frac{1/2}{3/8} = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$.Wait,re-evaluating the expansion: $\frac{n(n-1)}{2} = \frac{(-1/5)(-6/5)}{2} = \frac{6/25}{2} = \frac{3}{25}$.Coefficient of $x^2$ is $\frac{3}{25} \cdot (\frac{5}{2})^2 = \frac{3}{25} \cdot \frac{25}{4} = \frac{3}{4}$.So $a_2 = \frac{3}{4} \cdot 2^{-1/5}$.Then $\frac{a_1}{a_2} = \frac{1/2}{3/4} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$.

If $(2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + \ldots$,then $\frac{a_1}{a_2} = $

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